University Physics I · unit UPI-03
2D & 3D Kinematics
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Vector Position, Velocity & Acceleration: infer an input with uncertainty
The position vector r⃗(t) locates a particle from a chosen origin at each instant; its first time derivative is the velocity v⃗, tangent to the path, and its second is the acceleration a⃗, both taken component by component. Use the supplied topic model Y_out = abs(deg(atan(amp*(cos(t+dt)-cos(t))/dt)-atan(-amp*sin(t)))), where Y_out is CHORD OFF TANGENT (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; deg converts radians to degrees; unmarked trigonometric functions use radians. amp is Amplitude A, measured in m. Keep t (Time t) = 2 s; dt (Averaging interval Δt) = 1.5 s fixed. For this exercise use only 0.39125 m ≤ amp ≤ 1.71425 m on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 24.884823 °, with a bounded reading uncertainty ±0.25451 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown amp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001323 m, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
CHORD OFF TANGENT against amp
Predictions from the supplied Vector Position, Velocity & Acceleration model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CHORD OFF TANGENT / °
Horizontal axis (x): amp: Amplitude A / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| amp: Amplitude A / m | CHORD OFF TANGENT / ° |
|---|---|
| 0.39125 | 11.855027951701976 |
| 0.722 | 19.22530748313177 |
| 1.05275 | 23.688419511201182 |
| 1.3835 | 25.882350720955355 |
| 1.7142499999999998 | 26.581557928058707 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(amp) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.39125) = 11.855 ° and Y_out(1.71425) = 26.5816 °. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives amp_line = 0.39125 + (24.8848 − 11.855)(1.71425 − 0.39125)/(26.5816 − 11.855) = 1.56182 m. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives amp = 1.19828 m; substituting back gives Y_out = 24.884823 °.
- The interpolation error in the input is |amp_line − amp| = 0.363539 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 24.630313 ° and 25.139333 °. Sort the roots even if the response decreases: 1.16286 m ≤ amp ≤ 1.2372 m.
- The lower and upper input deviations are 0.0354162 m and 0.0389158 m. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer amp ≈ 1.19828 m; compatible interval [1.16286, 1.2372] m. Endpoint interpolation gives 1.56182 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Projectile motion: infer an input with uncertainty
A projectile is an object moving under gravity alone after launch: its acceleration is a constant 9.81 m s⁻² straight down, so its horizontal velocity stays fixed, its vertical velocity changes steadily, and its path is a parabola. Use the supplied topic model Y_out = v*cosd(th)*(v*sind(th)+sqrt(v^2*sind(th)^2+2*g*h))/g, where Y_out is RANGE AT θ (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; sind/cosd/tand use degrees; g = 9.81 m s⁻². th is Launch angle θ, measured in °. Keep v (Launch speed v₀) = 15 m s⁻¹; h (Launch height h) = 0 m fixed. For this exercise use only 17.4375 ° ≤ th ≤ 43.1625 ° on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 20.994914 m, with a bounded reading uncertainty ±0.28406 m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0025725 °, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
RANGE AT θ against th
Predictions from the supplied Projectile motion model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RANGE AT θ / m
Horizontal axis (x): th: Launch angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Launch angle θ / ° | RANGE AT θ / m |
|---|---|
| 17.4375 | 13.114402757514041 |
| 23.86875 | 16.97411518784345 |
| 30.3 | 19.981968144958472 |
| 36.73125 | 21.987010135529466 |
| 43.1625 | 22.888616532295526 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(th) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(17.4375) = 13.1144 m and Y_out(43.1625) = 22.8886 m. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th_line = 17.4375 + (20.9949 − 13.1144)(43.1625 − 17.4375)/(22.8886 − 13.1144) = 38.1784 °. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives th = 33.1298 °; substituting back gives Y_out = 20.994914 m.
- The interpolation error in the input is |th_line − th| = 5.04867 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 20.710854 m and 21.278974 m. Sort the roots even if the response decreases: 32.2772 ° ≤ th ≤ 34.0444 °.
- The lower and upper input deviations are 0.852577 ° and 0.914645 °. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer th ≈ 33.1298 °; compatible interval [32.2772, 34.0444] °. Endpoint interpolation gives 38.1784 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Relative Position & Velocity: infer an input with uncertainty
The velocity of P relative to A, written v_PA, is the rate of change of P's position measured by an observer at rest in A; for frames that do not rotate, v_PB = v_PA + v_AB links what any two frames measure. Use the supplied topic model Y_out = hypot(vb*cosd(phi),(vb*sind(phi)-va)), where Y_out is RELATIVE SPEED v_BA (m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²); sind/cosd/tand use degrees. va is A's speed due north, v_A, measured in m s⁻¹. Keep vb (B's speed, v_B) = 8 m s⁻¹; phi (B's heading φ, north of west) = 20 ° fixed. For this exercise use only 2.85 m s⁻¹ ≤ va ≤ 9.15 m s⁻¹ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 8.4952926 m s⁻¹, with a bounded reading uncertainty ±0.14653 m s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown va by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 m s⁻¹, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
RELATIVE SPEED v_BA against va
Predictions from the supplied Relative Position & Velocity model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RELATIVE SPEED v_BA / m s⁻¹
Horizontal axis (x): va: A's speed due north, v_A / m s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| va: A's speed due north, v_A / m s⁻¹ | RELATIVE SPEED v_BA / m s⁻¹ |
|---|---|
| 2.85 | 7.518402853289356 |
| 4.425 | 7.7049074525618195 |
| 6 | 8.1954906040295 |
| 7.574999999999999 | 8.940233980658949 |
| 9.149999999999999 | 9.881839455138001 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(va) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(2.85) = 7.5184 m s⁻¹ and Y_out(9.15) = 9.88184 m s⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives va_line = 2.85 + (8.49529 − 7.5184)(9.15 − 2.85)/(9.88184 − 7.5184) = 5.45401 m s⁻¹. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives va = 6.693 m s⁻¹; substituting back gives Y_out = 8.4952926 m s⁻¹.
- The interpolation error in the input is |va_line − va| = 1.23899 m s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.3487626 m s⁻¹ and 8.6418226 m s⁻¹. Sort the roots even if the response decreases: 6.36775 m s⁻¹ ≤ va ≤ 6.99852 m s⁻¹.
- The lower and upper input deviations are 0.325253 m s⁻¹ and 0.305517 m s⁻¹. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer va ≈ 6.693 m s⁻¹; compatible interval [6.36775, 6.99852] m s⁻¹. Endpoint interpolation gives 5.45401 m s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Uniform Circular Motion: infer an input with uncertainty
Uniform circular motion is motion on a circle of fixed radius at constant speed. The velocity keeps turning, so the particle accelerates toward the centre at a_c = v²/R even though its speed never changes. Use the supplied topic model Y_out = 2*pi/w, where Y_out is PERIOD T = 2π/ω (s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Angular speed ω, measured in rad s⁻¹. Keep r (Radius R) = 1.6 m; t (Time t) = 0.8 s fixed. For this exercise use only 0.659375 rad s⁻¹ ≤ w ≤ 1.84062 rad s⁻¹ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 4.553239 s, with a bounded reading uncertainty ±0.17094 s. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000118125 rad s⁻¹, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
PERIOD T = 2π/ω against w
Predictions from the supplied Uniform Circular Motion model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): PERIOD T = 2π/ω / s
Horizontal axis (x): w: Angular speed ω / rad s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Angular speed ω / rad s⁻¹ | PERIOD T = 2π/ω / s |
|---|---|
| 0.659375 | 9.529001413732074 |
| 0.9546875 | 6.581405231742938 |
| 1.25 | 5.026548245743669 |
| 1.5453125 | 4.065964202826021 |
| 1.840625 | 3.413615107465989 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(w) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.659375) = 9.529 s and Y_out(1.84062) = 3.41362 s. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.659375 + (4.55324 − 9.529)(1.84062 − 0.659375)/(3.41362 − 9.529) = 1.62049 rad s⁻¹. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives w = 1.37994 rad s⁻¹; substituting back gives Y_out = 4.553239 s.
- The interpolation error in the input is |w_line − w| = 0.240557 rad s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.382299 s and 4.724179 s. Sort the roots even if the response decreases: 1.33001 rad s⁻¹ ≤ w ≤ 1.43376 rad s⁻¹.
- The lower and upper input deviations are 0.0499317 rad s⁻¹ and 0.0538271 rad s⁻¹. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer w ≈ 1.37994 rad s⁻¹; compatible interval [1.33001, 1.43376] rad s⁻¹. Endpoint interpolation gives 1.62049 rad s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.